Matrices and Systems of Equations Help
Introduction to Matrices and Systems of Equations
There are three ways we can use matrices to solve a system of linear equations. Two of them will be discussed in this Study Guide. Solving systems using these methods will be very much like finding inverses. We will begin with 2 × 2 systems (two equations and two variables) and an augmented matrix of the form . A, B, C , and D are the coefficients of x and y in the equations and E and F are the constant terms. We will use the same steps above to change this matrix to one of the form . The numbers in the last column will be the solution.
The coefficients 2, −3, −1, and 1 are the entries in the left side of the matrix.
The constant terms 17 and −7 are the entries on the right side of the matrix.
The augmented matrix is
Step 1 We want −1, the C entry, to be 0.
Step 2 We want −3, the B entry, to be 0.
This row represents the equation −1x + 0 y = −4
This row represents the equation 0 x + (−1) y = 3
Step 4 Divide Row 1 by −1.
Step 5 Divide Row 2 by −1.
This row represents the equation 1 x + 0 y = 4.
This row represents the equation 0 x + 1 y = −3.
The solution is x = 4 and y = −3.
Begin solving a 3 × 3 system (three equations and three variables) by writing the augmented matrix
Using the same steps we used to find the inverse of a matrix, we want to change this matrix to one of the form .
The numbers in the fourth column will be the solution.
The augmented matrix is
Step 1 Use Row 1 and Row 2 to change the D entry to 0.
Step 2 Use Row 1 and Row 3 to change the G entry to 0.
Step 4 Because the B entry is already 0, this step is not necessary. New Row 1 is old Row 1.
Step 5 Use Row 2 and Row 3 to change the H entry to 0.
Step 7 Use Row 1 and Row 3 to make the C entry a 0.
Step 8 Use Row 2 and Row 3 to make the F entry a 0.
Step 10 Divide Row 1 by 8, Row 2 by 4, and Row 3 by −8.
The solution is x = −3, y = 4, and z = 2.
Finding the Inverse of a Matrix and Multiplying Two Matrices
The second method we will use to solve systems of equations involves finding the inverse of a matrix and multiplying two matrices. We begin by creating the coefficient matrix and the constant matrix for the system.
The coefficient matrix is and the constant matrix is . We will find the inverse of the coefficient matrix and multiply the inverse by the constant matrix. The product matrix will consist of one column of two numbers. These two numbers will be the solution to the system.
The coefficient matrix and constant matrix are
The next matrix is . We need to divide Row 1 and Row 2 by − 5.
Multiply the inverse matrix by the coefficient matrix.
The solution is x = 4 and y = 1.
The strategy is the same for a 3 × 3 system of equations.
The coefficient matrix and the constant matrix are
We will find the inverse matrix of the coefficient matrix and multiply it by the constant matrix.
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